COVENANT UNIVERSITY CANAANLAND, OTA

OMEGA SEMESTER 2007/2008 SESSION

COURSE TITLE: GENERAL ORGANIC CHEMISTRY

COURSE CODE: CHM 121

LECTURER: AJANI O.O.

Topics under This Section are:

(1) Hybridization.(2) Formation of sp3, sp2, sp orbitals in carbon.(3) Homologous series.(4) Functional groups.(5) Isomerisim.

Objectives: At the end of this section of the course, students are expected to know the following:

What is hybridization? Hybridization in carbon atom systems Homologous series: introduction and characteristics. Various classes of functional groups. General and selective tests for some functional groups. Some carboxylic acid derivatives. Isomerism - both structural and stereoisomerism. Ordinary and plane polarized light.

Hybridization: can be defined as the blending of two or more atomic orbitals of different energy to obtain a hybrid orbital of equivalent energy level. This depends on the concept of overlapping of atomic orbital and mathematical summing of their wave functions towards a specific direction. Every overlap of atomic orbitals results in bond formation which could be either sigma (δ) or pi (π) bond. However, the type of bond formed depends on the nature of the overlap e.g. axial overlap results in sigma (δ) bond while lateral (side-to-side) overlap results in pi (π) bond formation. This is why sigma (δ) bond is stronger than pi (π) bond. Hybridization helps us to know the molecular geometry of the compounds, their bond angles, the nature of electron pair [i.e. number of Bond Pair (BP), Lone Pair (LP)]. When all the electron

1

pairs surrounding the central atom are bond pair (BP), the compound shows the normal shape peculiar to its hybridization pattern. However, when there is lone pair (LP), there will be slight distortion to give it another shape because the lone pair (LP) tends to re-orientate itself in order to minimize repulsion. The order of the repulsive power is: LP-LP > LP-BP > BP-BP. This is established by Valence Shell Electron Pair Repulsion Theory (VSEPAR).

Determination of Hybridization Pattern Write out the ground state (GS) valence electronic configuration of central atom. Promote the excited state (ES) in such a way to accommodate electrons from

surrounding groups or atoms. Pair each electron from the surrounding atoms with each half-filled suborbital level of

the central atom to have fully filled configuration at each suborbital level. Write out the valence suborbital level at this stage as the hybrid orbital. You can then predict the shape, bond angle and some other factors as shown in the

table below:

Total Electron Pair

No of Bond Pair (BP)

No of Lone Pair (LP)

Hybrid orbital

Geometrical Distribution or Shape

Ideal Bond Angle

Typical Example

2 2 0 sp Linear 180O BeCl2

33

32

01

sp2

sp2Trig. PlanarTrig. Planar

120O

120OBF3

SiO2

444

432

012

sp3

sp3

sp3

TetrahedralTrig. PyramidAngular/Bent

109.5o

107 o

104.5 o

CH4

NH3

H2O

5555

5432

0123

sp3dsp3dsp3dsp3d

Trig. BipyramidTrig. PyramidTrig. PlanarLinear

90o, 120o

90o, 120o

120o

180o

PCl5

SF4

ClF3

XeF2

666

654

012

sp3d2

sp3d2

sp3d2

OctahedralSq. pyramidSq. planar

90o

90o

90o

SF6

BrF5

XeF4

Example: What is the hybridization pattern or the hybrid orbital of Sulphur tetrafluoride (SF4) and use it to predict the shape and bond angle of the compound.

2

Solution:

(i) Central atom is Sulphur (16S) with G.S. electronic configuration of 1s2 2s2 2p6 3s2 3p4

G.S. valence electronic configuration

(ii) Ex. State valence electronic conf.

(iii) Each of the up spin half filled sub-orbital accepts each of the 4down spin electron

from fluorine

(iv) Hybrid. Pattern of SF4 = sp3d (with 4BP + 1LP = 5TP).

(v) Shape is Trig. Pyramid with bond angle of 90o and 120o

Hybridization in Carbon

Considering the carbon context, hybridization is the blending or mixing of s and p orbitals in the valence shell to produce equivalent orbital known as Hybrid orbitals. There are three main types of hybridization in carbon: They are sp3, sp2 and sp. The hybrid orbitals are named after the atomic orbitals from which they were constructed or formed e.g.

2s + 2p3 Four of sp3 hybrid orbitals.

2s + 2p2 Three of sp2 hybrid orbitals.

2s + 2p Two of sp hybrid orbitals.

3

Sp3 Hybridization in Carbon

2s orbital mixed with three 2p (i.e. 2px, 2py and 2pz) orbitals to form four sp3 hybrid orbitals of equivalent energy which are tetrahedrally arranged (to minimize repulsion) with the bond angle (B.A.) of 109.5o. Carbon will be sp3-hybridized when it is directly bonded to four atoms in a neutral molecule e.g.

Orbital Blending Diagram

A Typical sp3 Hybridization in Methane (CH4)

Each of the half-filled four sp3 hybrid orbital can use bigger side of its lode to overlap with each 1s orbital of hydrogen to form 4 bonds in methane.

4

Sp2 Hybridization in Carbon

2S orbital mixed with two of its 2p (i.e. 2px, 2py) orbitals to obtain three sp2 hybrid orbitals of equivalent energy which are all in the same plane with trigonal planar arrangement having bond angle of 120o. Here, the third 2p orbital (i.e. 2pz) is unhybridized. Carbon will be sp2-hybridized if it is directly bonded to three atoms in a neutral molecule e.g.

Orbital Blending Diagram:

Sp Hybridization in Carbon

The carbon 2s orbital combines with one of its 2p orbital (i.e. 2px) to generate two sp hybrid orbitals of equivalent energy which are linear in shape with bond angle of 180o. The remaining two 2p (i.e. 2py, 2pz) are unhybridized. This sp hybrid orbital has 50% s and 50%p character. Carbon will be sp-hybridized when it is directly linked to two atoms in a neutral molecule e.g.

5

Orbital Blending Diagram:

Formation of Sigma (δ) and Pi (π) Bonds

Having seen the hybridization patterns (sp3, sp2 and sp) of a single carbon, it is of great interest to know that this carbon can overlap with another carbon to form either sigma (δ) or pi (π) bond. The hybrid orbital of adjacent carbon atoms can overlap with one another or with atomic orbital of other elements such as sulphur, phosphorus, oxygen, nitrogen etc to form covalent bond which could either be (δ) or (π) bond.

Sigma bond can be formed by any of the following three overlaps:

(i) Overlap of two hybrid orbitals: These include sp3-sp3, sp2-sp2, sp-sp, sp3-sp2, sp3-sp, sp2-sp. For instance, the kind of overlap seen in carbon-carbon bond of ethane is sp3-sp3

(ii) Overlap of a hybrid orbital and s orbital e.g. sp3-s, sp2-s, sp-s. This is seen in carbon-hydrogen bone of methane and other alkanes.

6

(iii) Overlap of a hybrid orbital and p orbital that is linearly aligned e.g. sp3-p, sp2-p, sp-p. For example sp3-p is seen in overlap of carbon with p-block elements such as chlorine.

Pi –bond: can only be formed when two adjacent carbons are unhybridized. It is a covalent bond formed as a result of lateral overlap of two unhybridized p-orbitals on adjacent carbon atoms.

Sp3-sp3 Overlap in Ethane (C2H6)

Two sp3 hybridized carbon atoms formed carbon-carbon sigma bond in ethane as shown below both in orbital picture and 3D-stucture.

In summary ethane contains 1δ of C-C + 6δ of C-H = 7δ bonds in all.

7

Sp2-sp2 Overlap in Ethene (C2H4)

When two sp2 hybridized carbon atoms link up, they form a double bond (C=C) which is made up of 1δ and 1π-bonds. The δ-bond is formed as a result of overlap of the hybridized orbital sp2-sp2 while π-bond is formed as a result of lateral overlap of unhybridized 2pz orbital of one carbon with that of the other as seen in ethene. In summary ethane has 1δ of C-C + 4δ of C-H + 1πbond of C-C = 5δ + 1π bonds.

The electrons in the π-bond are called π electrons while those in the δ-bond are called δ electrons. Since π-bond is weaker than δ-bond, the bond energy of C=C is less than twice the bond energy of C-C.

Rotation about C=C is restricted unlike that involve in C-C which is free. The bond length is 1.34Å (shorter because of higher electron density).

Sp-sp Overlap in Ethyne (C2H2)

Overlap of sp-sp hybrid orbital of two carbon atoms gives rise to a triple bond (C Ξ C), which has 1δ +2π-bonds. The 2π-bonds are product of lateral overlap of two unhybridized p orbitals (i.e. 2py, 2pz) of one sp carbon with that of other as seen in ethyne (C2H2). The whole of ethyne in summary, has 3δ + 2π bonds.

8

Rotation about C Ξ C is the most difficult. The bond energy of a C Ξ C is also less than thrice the bond energy of C-C, it has the shortest bond length of 1.20 Å because of its highest electron density.

Example 2: Assuming the normal bond angle at sp3, sp2 and sp hybridized carbon atom, draw the graphical formula of each of the following compounds in 3-Dimensional representation.

(i) CH3CH2 C Ξ CCH=CH2 (Hex-1-en-3-yne)

(ii) H2C=C(CH3)CH=CH2 (2-Methylbuta-1,3-diene or isoprene)

(iii) CH3CH=CHCH2 C Ξ CH (Hex-4-en-1-yne)

(iv) CH3CH=CHCH3 (Both cis- and trans-but-2-ene)

(v) CH3 C Ξ CCH=CH C Ξ CH (Hept-3-en-1,5-diyne).

Solution2: To solve this problem, you need 3 things.

Identification of the type of hybridization for each carbon.

Draw the graphical formula showing the bond angles and shape (geometry).

Put each bond angle at appropriate carbon center.

Questions for Practice.

1. Draw the orbital pictures of all the compounds (i) to (v) in the example 2 above.

9

2. What are the two types of covalent bonds; hence, which of the two is stronger and why?

3. Using the common hybridization pattern, what is the hybrid orbital of BF3 and state why it is being referred to as a Lewis acid.

4. Each of CH4 and NH3 has a total of 4 electron pairs surrounding the central atom. However, CH4 is tetrahedral while NH3 is trigonal pyramid. How can you account for this using hybridization pattern?

5. Give two examples of compounds in each , in which the following overlaps occur:

(i) Sp3-sp3 overlap (iv) sp2-p overlap

(ii) Sp3-sp2 (v) sp3-d overlap

(iii) Sp-p overlap (vi) sp-s overlap

Note: in each case, show the resulting bond of such overlap.

Homologous Series and Functional Groups

Compounds that are member of the same family (i.e. compounds having the same functional groups) but differ from their neighbours in their molecular formula by a constant unit –CH2- are called members of a homologous series. For example, alkane is a family of saturated hydrocarbon with its first four members being methane (CH4), ethane (C2H6), propane (C3H8) and butane (C4H10). Each of these hydrocarbons differs from its neighbor by a constant of –CH2- despite the fact that they are all alkanes; so, they are members of the same homologous series.

Characteristics of a Homologous Series

(1). All members can be represented by a general formula e.g. the general formula of alkane is CnH2n+2

When n = 1, we have CH4 which is methane

When n = 2, we have C2H6 which is ethane

When n = 3, we have C3H8 which is propane

When n = 4, we have C4H10 which is butane.

The prefix “alk-“ is determined by the number of carbon atom present while the

suffix “-ane” implies the presence of saturation, e.g. C1(meth-), C2(eth-), C3(prop-),

10

C4(but-), C5(pent-), C6(hex-), C7(hept-), C8(oct-), C9(non-), C10(dec-)……………….

and C20(eico-).

(2). Each member differs from the next by a constant of –CH2-, e.g.

C4H10 – C3H8 = C3H8 – C2H6 = C2H6 – CH4 = -CH2-

(3). Members can be prepared by the same method. For instance, alkanes can be prepared by the action of sodium alkanoate on NaOH. For the preparation of methane (CH4), sodium ethanoate (CH3COONa) is used while for the preparation of ethane C2H6, sodium propanoate (CH3CH2COONa) is the most appropriate precursor. However, the rate of reaction will be different. The bulkier the alkyl substituent the lower the rate of reaction. Some reaction will not even proceed without the presence of a catalyst.

CH3COONa + NaOH CH4 + Na2CO3

CH3CH2COONa + NaOH C2H6 + Na2CO3

(4). Members of the same homologous series have similar chemical properties e.g. all alkanes undergo combustion reaction to give CO2 and H2O irrespective of their state of matter.

CH4(g) + 2O2 CO2 + 2H2O

C7H16(l) + 11O2 7CO2 + 8H2O

C17H36(s) + 26O2 17CO2 + 18H2O

(5). Physical properties of the members vary with change in the molecular weight.

C1 – C4 are gases at room temperature.

C5 – C14 are liquid at room temperature.

C15 are above are solid at room temperature.

1. Write out the structural formular of methanoic acid (formic acid), ethanioc acid (acetic acid) and propanoic acid (propionic acid) showing all the necessary bonds. Hence, or otherwise In not more than three sentences, how can you establish that they are members of the same family?.

11

2. Nonadecane (C19H40) is a solid hydrocarbon with the properties of a wax; write out a balanced chemical equation for the complete combustion of this alkane in air. How can you test for each of the products of this reaction?

3. Write a balanced chemical equation for the reaction of sodium pentanoate with sodium hydroxide under acid catalyzed medium. What is the relevance of the product of this reaction?.

4. Decane used up x moles of oxygen in a complete combustion to produce y moles of carbon dioxide and z moles of water. Using a balanced chemical equation, determine the value of x, y and z. Hence, solve these equations:

(i) 2x2 + 3y2 + z2 – 5xyz. (iii) 4x2 + (2y)2 – 2(xy + yz)

(ii) 3x + 3y – 3z + xyz (iv) x + y - 21zo

Functional Group: can be defined as an active part of a compound or molecule that takes part in chemical reaction. It depicts the family of an organic compound and at the same time determines its properties. Functional group is responsible for the type of reaction a compound undergoes under a prescribed set of condition. All chemical reactions in the whole world are just conversion of functional groups from one form to the other.

Reaction Mechanism: is the pathway or route that explains how the structure of the reactants is transformed to that of products.

S/N IUPAC(Trivial) Name Suffix General Form.

Functional Group

Typical Example

1 Alkane -ane CnH2n+2 -C- paraffin Methane

2 Alkene -ene CnH2n C=C Ethene

3 Alkyne -yne CnH2n-2 C Ξ C Propyne

4 Alkoxy alkane -ether R-O-R’ -O- (ether) Diethylether

5 Alkanol (Alcohol) -ol R-OH -OH (hydroxyl) Ethanol (1o)

6 Alkanal (aldehyde) -al R-CHO C=O (carbony of aldehyde)

ethanal

7 Alkanone (ketone) -one R-CO-R’ C=O (carbonyl Propanone

12

of ketone) (acetone)

8 Akanoic acid (carboxylic acid)

-oic acid

R-COOH -COOH (carboxylic)

Ethanoic acid CH3COOH

9 Amine -amine R-NH2 -NH2 (amino) 1o, 2o or 3o

Ethanamine CH3CH2NH2

Test for Alkane, Alkene and Alkyne

Alkene and alkyne will undergo a positive reaction and decolourize the following reagents because of the presence of unsaturation whereas alkane will not react with them.

Acidified potassium permanganate ( KMnO4/ H+).

Alkaline potassium permanganate ( KMnO4/ HO-).

Acidified potassium heptaoxochromate vi (K2Cr2O7/ H+).

Bromine water in carbon tetrachloride (Br2/H2O IN CCl4).

Hydrogen gas in finely divided Nickel catalyst (H2/Ni at 200oC).

However, to distinguish between alkene and alkyne, ammoniacal silver nitrate or ammoniacal cupper oxide is employed. While alkene remains unreactive in ammoniacal solution, terminal alkyne will react positively with such solution.

CH2=CH2 + Br2/H2O HO-CH2-CH2-Br

CH2=CH2 + KMnO4/ HO- HO-CH2-CH2-OH

Test for Carbonyl Group

There are two kinds of carbonyl; they are (i) carbonyl of aldehyde and (ii) carbonyl of ketone.

Carbonyl of aldehyde is found at the terminal end of a chain while that of ketone is inserted between two carbon atoms.

13

So, in aldehyde, one hydrogen is attached to the carbonyl carbon while in ketone the H had been replaced with carbon.

(a).DNP Test (General Test for Carbonyl)

Due to the close relationship exhibited by the two, they both react positively with 2,4-dinitrophenyl hydrazine (2,4-DNP) to give colored crystalline solid called 2,4-dinitrophenyl hydrazone .

Since, both aldehye and ketone have the same reaction with 2,4-DNP how then can we distinguish between the two? This can be achieved by using either Fehling or Tollen’s test; since aldehyde react positively with the two reagents (Fehling and Tollen’s reagents) while ketone will not. Otherwise, you can react ketone with semicarbazide (NH2NHCONH2) to obtain a coloured precipitate of semicarbazone while aldehyde will not react with semicarbazide.

(b).Fehling Solution Test

Add few drops of alkanal (aldehyde) to the blue coloured fehling solution and warm; the mixture will turn to brick red precipitate whereas with alkanone (ketone) the blue colour remains (i.e. no reaction).

This change is as a result of reduction of Cu+2 to Cu+1

RCHO + Cu+2 NaOH/H2O/warm RCOONa + Cu+1

Blue brick red

This is also what happens in test for glucose because it is a reducing sugar. Glucose will give a positive reaction with fehling solution while fructose will not. Fehling solution is a mixture of fehling A (CuSO4) and Fehling B (NaOH and Rochelle salt).

14

(c).Tollen’s Test

This is otherwise called silver mirror test. Warming a mixture of aldehyde and ammoniacal silver nitrate (Tollen’s reagent) gives a deposit of shining silver metal on the wall of the test-tube whereas with carbonyl of ketone, there is no reaction.

RCHO + Ag(NH3)2+ RCOO- + Ag

Tollen’s reagent (AgNO3/NH3) is good for the synthesis of α,β-unsaturated acid from α,β-unsaturated aldehydes because it does not attack the double bond (C=C) in the process.

R-CH=CH-CHO Ag(NH3)2+ R-CH=CH-COOH

α,β-unsaturated aldehyde α,β-unsaturated acid

Test for Phenol

Add few drops of freshly prepared iron III chloride (FeCl3) to the phenol and shake lightly; observation of an intense colouration confirms the presence of phenolic –OH. This test is very good for the confirmation of the polyphenolic groups (anthocyanin) in natural product.

Test for -OH of Alcohol

There are three types of alcohols. These are the primary (1o), the secondary (2o) and the tertiary (3o) alcohols. All these alcohols give steamy or dense white fume when treated with PCl5 or PCl3 or thionyl chloride at room temperature.

ROH + PCl5 RCl + POCl3 + HCl

However to distinguish between 1o, 2o and 3o alcohol, Lucas reagent is being used. Add few drops of Lucas reagent to the alcohol; with 1o alcohol, there is formation of a clear liquid; with 2o, the liquid separates into two layers after 5minutes while with 3o alcohol, the separation into two layers happens immediately.

Lucas reagent is prepared by dissolving 0.5 mole of ZnCl2 in 0.5 mole of concentrated HCl followed by cooling at room temperature.

Test for Carboxylic Acid (-COOH)

15

When few drops of saturated sodium bicarbonate (NaHCO3) is added to any carboxylic acid group, there is an effervescence of a gas (CO2) which turns lime water milky. However, the milkiness disappears when the gas is liberated in excess.

RCOOH + NaHCO3 RCOONa + H2O + CO2

Preparation of Carboxylic Acid Derivatives

Carboxylic acid derivatives are the compounds in which the –OH of carboxyl functional group has been replaced by another element or group.

1. Acid Halide: This is obtained when the –OH of COOH is being replaced with a halogen (e.g. Y = F, Cl, Br, I). Acid chloride is much more reactive than the carboxylic acid from which it is being obtained. It can be prepared by treating carboxylic acid with either thionyl chloride (SOCl2) or PCl5.

(ii) CH3COOH + PCl5 CH3COCl + POCl3(l) + HCl(g)

(iii) CH3COOH + SOCl2 CH3COCl + SO2(g) + HCl(g)

2. Amide: This is obtained when the –OH of COOH is being replaced with an amino group (e.g. Y = NH2). This is a derivative of carboxylic acid that is formed by the reaction of ammonia or amine with acid halide or carboxylic acid.

CH3COOH + NH3 CH3CONH2 + H2O

16

CH3COCl + NaNH2 CH3CONH2 + NaCl

Ethanoyl chloride ethanamide

3. Acid Anhydride: This is obtained when the –OH of COOH is being replaced with an alkoxylcarbonyl group (e.g. Y = OCOCH3). This is a derivative of carboxylic acid that is formed by distillation of two molecules of the acid in the presence of a suitable dehydrating agent such as P4O10. The driving force for the reaction is the elimination of a stable molecule of water.

4. Ester: This is obtained when the –OH of COOH is being replaced with an alkoxy group (e.g. Y = OR). Ester can be obtained by the reaction of the acid with alcohol in the presence of a mineral acid. However, if the acid chloride is used the reaction takes place at low temperature (0-5oC ) and need no catalyst. This forward reaction is called esterification (alkanoation) while the backward reaction is called hydrolysis.

CH3COOH + CH3CH2OH H+ CH3COOCH2CH3 + H2O

Ethyl ethanoate

CH3COCl + CH3CH2OH 0-5oC CH3COOCH2CH3 + HCl

Questions for Practice

1. An unlabelled bottle on the shelf is suspected to contain either propanal, propanone or propanoic acid. Describe the chemical test(s) to establish which of the three is actually in the bottle.

2. Name four classes of carboxylic acid derivatives with two examples of each. Then, explain how each of those compounds can be obtained from the corresponding carboxylic acid.

17

3. The compound Y below is the structure of the ampicillin which is a β-lactam antibiotic used in the treatment of infections. Circle and write the name of the functional group(s) present in Y. Hence, explain briefly how to test for those identified functional group(s).

4. With the aid of equation only, how can you synthesize 3-methyl-2-pentenoic acid from 3-methyl-2-pentenal .

Isomerism

In 1823, transformation of ammonium cyanate (inorganic compound) to urea (organic compound) was first observed by Wohler who was a student of J.J. Berzelius in Germany. Then, Berzelius was interested in a case in which two clearly different materials had the same elemental composition and he invested the term ”isomerism” to describe it.

Isomerism: is said to occur when two or more compounds have the same molecular formula but different physical properties and/or chemical properties. The compounds that exhibit this behavior are called Isomers. There are two main types of isomerism: (i) structural (ii) stereoisomerism.

18

1. Structural Isomerism: This occurs when two or more compounds have the same molecular formula but different structure (i.e. arranging atoms in different ways). It is subdivided into three namely: positional, functional group and chain isomerism.

(i). Positional Isomerism: This occurs within a particular homologous series that involves changing of the position of some substituents. This type of isomerism has the same chemical properties but different physical properties.

(ii). Functional Group Isomerism: This occurs when two or more isomers have the same molecular formula but different functional groups, e.g. alcohol and ether.

Presence of polar –OH with formation of Hydrogen bonding makes alcohol to have high boiling point while low boiling point and high volatility in ether is as a result of the presence of weak force of attraction. Another example of the functional group isomers are alkanal (aldehyde) and alkanone (ketone).

19

(ii). Chain Isomerism: This occurs when two or more isomers have their chains differently arranged. It occurs within the same homologous series. There is an occurrence of straight and branched chain isomers. However, the straight chain isomers have higher boiling point and lower volatility than the branched chain counterparts despite they have the same molecular formula e.g.

2. Stereoisomerism: This occurs when two or more compounds have the same molecular formula but different arrangement in space. This is established by using the three dimensional structure of sp3, sp2 and sp hybridization of carbon atoms. It is subdivided into two namely: geometrical and optical isomerism.

(i). Geometrical Isomerism: This exists in compounds where free rotation is not possible either because of C=C or presence of ring in cyclic compounds. For this to occur C=C must be symmetrically positioned. It is exhibited inform of cis- and trans-isomers. The two atoms attached to a sp2 carbon must not be the same in order to have cis-trans isomerism, e.g.

Another example of the compound that exhibits geometrical isomerism is 1,2-dichloroethene. ClCH=CHCl.

20

(ii). Optical Isomerism: This occurs when a compound is able to rotate a plane polarized light. When a plane polarized light is made to pass through the solution of an optically active compound, the plane of polarization rotates to the right or clockwisely to give a dextrorotatory, (+) compound. If the rotation is to the left or anticlockwise in nature, it gives lavorotatory compound which is indicated with the symbol (-) as the prefix. For any isomer to be optically active, it must possess an asymmetric carbon atom (i.e. carbon with a stereogenic or chiral center). Sp3 carbon atom is said to have a chiral center if it is surrounded by four different atoms or groups. If a compound has a chiral carbon center but it is not superimposable on its mirror image, it is called an enantiomer while the phenomenon is known as enantiomerism. Enantiomeric pairs have the same physical properties, except for the direction of rotation of plane polarized light, e.g. 1-chloropropan-1-ol and 2-hydroxypropanoic acid (lactic acid) have one chiral carbon center each, so, they are optically active and their enantiomeric pair will be the mirror image that is not super imposable on the starting isomer.

21

Other examples of optical isomers are:

(iii) chloroiodomethane sulfonic acid (ICHCl.SO3H)

(iv) 2-bromobutane (CH3CH2CHBr.CH3)

Ordinary and Plane Polarized Light

Ordinary Light: can be defined as electromagnetic radiation vibrating at right angle to the direction of its propagation (transverse wave motion) in an infinite numbers of plane. When ordinary light passes through polarizer, its light vibrates only in one plane and become polarized. When this plane polarized light is made to pass through solution of some compounds, the plane of polarization is either rotated to the right or left.

22

23